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Let $M$ be a smooth $m$-manifold and $N_1$, $N_2$ smooth embedded $k$-submanifolds such that

  • $N_1\cap N_2=\partial N_1=\partial N_2$,
  • for each $x\in N_1\cap N_2$, $T_x N_1=T_x N_2$, and
  • for $x\in N_1\cap N_2$, each vector in $T_x N_1$ is inwards-pointing wrt. $N_1$ iff it is outwards pointing wrt. $N_2$ and vice versa.

Is it true that $N_1\cup N_2$ is a smooth embedded $k$-submanifold with $N_1\cap N_2$ in its interior?

Edit: It follows from mollyerin's answer that $N$ is not necessarily the image of a $C^\infty$ embedding. But I have further bounty questions:

  • Is it at least the image of a $C^1$ embedding?
  • Are there some further simple conditions that I should include so that $N$ is a smooth ($C^\infty$) embedded submanifold?
  • Can I "change" $N$ in some arbitrary small neighborhood of $N_1\cap N_2$ to make the resulting manifold smooth?
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  • $\begingroup$ Could you please comment on the downvote? $\endgroup$ Jan 23 '15 at 21:00
  • $\begingroup$ I didn't downvote, but there is a faction on this site who think it's not okay to ask a question unless you show some work that you've done. $\endgroup$ Jan 23 '15 at 21:05
  • $\begingroup$ Ok, right, thanks. I will work on the quesiton. $\endgroup$ Jan 23 '15 at 21:05
  • $\begingroup$ Consider something like this: N1 be the upper branch of $y=x^3$, and N2 be the lower branch of $y=-x^3$. $\endgroup$
    – Xipan Xiao
    Jan 23 '15 at 23:48
  • $\begingroup$ @XipanXiao Right, thanks. Of course I should add that invards pointing vectors in $T_x N_1$ wrt. $N_1$ are outwards pointing wrt. $N_2$. Now is it sufficient? $\endgroup$ Jan 24 '15 at 0:03
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This is false; you'll need assumptions (I'm not sure what good ones are) that ensure that the manifolds glue together "smoothly". As a counterexample, let $f_1: (-\infty, 0] \to \mathbb{R}^2$ be the map $f_1(x) = -x^2$, and let $f_2 : [0, \infty) \to \mathbb{R}^2$ be given by $f_2(x) = x^2$. We of course let $N_1$ be the image of $f_1$ and $N_2$ the image of $f_2$. Then $N_1$ and $N_2$ satisfy your conditions. However, $N = N_1 \cup N_2$ is not a smooth embedded manifold.

[Proof: If it were, there would be a smooth embedding $g : \mathbb{R} \to \mathbb{R}^2$ whose image is $N$. Say $g = (g_1, g_2)$. The map $g_1$ satisfies $g_1'(t) \neq 0$ for all $t$, and so (since it is also smooth), it is a diffeomorphism $\mathbb{R} \to \mathbb{R}$ (by the inverse function theorem). Therefore the map $g \circ g_1^{-1}$ is a smooth embedding $\mathbb{R} \to \mathbb{R}^2$ whose image is $N$, and this map has the form $x \mapsto (x, g_2 \circ g_1^{-1}(x)$), so evidently $g_2 \circ g_1(x) = sign(x) x^2$. But this map is not smooth. ]

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  • $\begingroup$ Thanks. Can I at least say that $N$ is a $C^1$-submanifold, or something like that? $\endgroup$ Jan 24 '15 at 11:12
  • $\begingroup$ Seems like $N$ should be the image of a $C^1$ embedding of a smooth manifold, but the proof isn't immediately obvious to me (I get confused about these things); you might try unaccepting my answer and editing the original question to see if anyone has ideas or a reference. (I will also try thinking about it.) $\endgroup$
    – mollyerin
    Jan 24 '15 at 11:30
  • $\begingroup$ Thanks. I will leave the acceptance now and offer a bounty tomorrow. $\endgroup$ Jan 24 '15 at 12:52
  • $\begingroup$ I removed the acceptance, at least temporarily, because I'm afraid that otherwise I can't attract attention.. $\endgroup$ Jan 30 '15 at 10:49
  • $\begingroup$ @PeterFranek That seems wise :D $\endgroup$
    – mollyerin
    Jan 30 '15 at 11:04

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